# How would you balance the following equation: I2+HNO3-->HIO3+NO2+H2O?

Nov 13, 2015

This is a redox reaction, which may be approached by oxidation and reduction half equations.

#### Explanation:

Oxidation:
Elemental iodine is oxidized to $I {O}_{3}^{-}$, iodate.

$\frac{1}{2} {I}_{2} + 3 {H}_{2} O \rightarrow I {O}_{3}^{-} + 5 {e}^{-} + 6 {H}^{+}$ $\left(i\right)$

Reduction:

Nitric acid is reduced to nitric oxide ${N}^{V} \rightarrow {N}^{I V}$:

$H N {O}_{3} + {H}^{+} + {e}^{-} \rightarrow N {O}_{2} + {H}_{2} O$ $\left(i i\right)$

The overall redox reaction is $\left(i\right) + 5 \times \left(i i\right)$ $=$ the overall redox equation:

$\frac{1}{2} {I}_{2} + 5 H N {O}_{3} \rightarrow I {O}_{3}^{-} + 5 N {O}_{2} + 2 {H}_{2} O + {H}^{+}$

Alternatively, I could write:

$\frac{1}{2} {I}_{2} + 5 H N {O}_{3} \rightarrow H I {O}_{3} + 5 N {O}_{2} + 2 {H}_{2} O$

Are the final equations balanced with respect to mass and to charge? Don't trust my arithmetic. How would I remove the half coefficient on ${I}_{2}$?